The simplest account of spontaneous symmetry breaking goes like this.

  • Take a potential $V(\phi)$ with symmetric minima that are not at $\phi = 0$, like the Mexican hat potential shown in this site's logo.
  • Since variations in the field cost energy due to the $(\partial_\mu \phi)^2$ term, minimum energy configurations have constant $\phi$.
  • Therefore, the lowest energy states have $\phi$ equal to one of the minima of $V(\phi)$. Thus we have symmetry breaking, because the vacuum state (whichever one we choose) does not have the symmetry that $V$ had.
  • In the quantum case, everything works the same, except the classical solution $\phi = c$ becomes $\langle \phi \rangle = c$. Then we have multiple vacuum states, each of which break the symmetry.

I'm suspicious about the last assertion. Suppose $V$ has two mimima, giving two degenerate vacuum states, $|+\rangle$ and $|-\rangle$.

Quantum mechanics allows superposition, so can we not take $(|+\rangle + |-\rangle)/\sqrt{2}$ as our vacuum? This state does not break the symmetry at all.

  • $\begingroup$ But that state does not correspond to a classical minimum of the potential, and is hence unsuited for doing the usual perturbation theory around it. $\endgroup$
    – ACuriousMind
    Commented Mar 13, 2016 at 23:04
  • $\begingroup$ This seems to suggest that spontaneous symmetry breaking is not an actual physical thing, but something we impose for convenience. Is that right? $\endgroup$
    – knzhou
    Commented Mar 13, 2016 at 23:07
  • 1
    $\begingroup$ That depends on what you mean by "actual physical thing". SSB does not actually remove the symmetry, it just means that the symmetry is realized non-linearly on the "correct" dynamical variables for the QFT (which are perturbations around the classical minima) (see, for instance, section 2.8 of this and references therein). $\endgroup$
    – ACuriousMind
    Commented Mar 13, 2016 at 23:17
  • $\begingroup$ @ACuriousMind "SSB does not actually remove the symmetry" -- I would say this very much depends on the situation. E.g., for a Ising model, the actual ground state will break the symmetry, since only the symmetry-broken states are stable against fluctuations. For other scenarios (such as breaking of U(1) in the BCS state), such fluctuations might be ruled out on physical grounds, and the symmetry breaking of the ground state is just a convenient assumption. Of course, from a practical point of view, it doesn't make a difference. $\endgroup$ Commented Dec 29, 2016 at 13:23

1 Answer 1


The state that you are proposing is prohibited by something often called a superselection rule. There are at least two ways to look at it.

First of all, spontaneous symmetry breaking is only possible in infinite volume. One can show, that in the limit of large volume, any matrix element $$ \left\langle +\right\vert A\left\vert -\right\rangle $$ between different vacuum states tends to zero. This means that there is no way to evolve, excite, etc, from one vacuum state to another (essentially because it would take infinite energy). Therefore, the states built on one of the vacuum states do not "talk" to the states built on other vacuum states, and your full Hilbert space is a direct sum $$ \mathcal{H}=\mathcal{H}_+\oplus\mathcal{H}_-, $$ and none of the observables take you from $\mathcal{H}_+$ to $\mathcal{H}_-$ or vice versa. While you can consider the linear combinations you are suggesting, there is clearly no physical meaning to it.

Another point of view is that the true vacuum states $\left\vert \pm\right\rangle$ are the ones for which the cluster decomposition property holds, and it does not hold for the linear combinations. Cluster decomposition property is the very physical idea that well-separated experiments on the vacuum state should not be correlated. More precisely, it says $$ \left\langle \Omega\right\vert \phi(x_1)\ldots\phi(x_n)\phi(y_1+z)\ldots\phi(y_k+z)\left\vert \Omega\right\rangle\to\left\langle \Omega\right\vert \phi(x_1)\ldots\phi(x_n)\left\vert \Omega\right\rangle\left\langle \Omega\right\vert \phi(y_1)\ldots\phi(y_k)\left\vert \Omega\right\rangle, $$ for sufficiently large spacelike $z$. It is clear that if this property holds in some vacuum states, it does not hold in their arbitrary linear combinations, since the equation is non-linear in $\Omega$. You can actually treat it as the equation which determines the true ground states.

I recommend reading the chapter on spontaneous symmetry breaking in Weinberg QFT II.


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